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The speed of a projectile when it reaches its maximum height

  1. May 21, 2016 #1
    1. The problem statement, all variables and given/known data
    The speed of a projectile when it reaches its maximum height is one-half its speed when it is at half its maxi- mum height. What is the initial projection angle of the projectile?

    2. Relevant equations
    Vf2 = vi2 + 2ax


    3. The attempt at a solution
    I know that the speed at the top only consist of an x component and that the speed at half the height would be sqrt(vx2+vy2) . But im trying everything from two hours and not getting a solution. The answer is suppost to be 67.8 but I dont know how to get it someone help
     
  2. jcsd
  3. May 21, 2016 #2
    The max height of the projectile is given by [itex] \frac {u^2sin^2(\theta)}{2g} [/itex] where [itex] \theta [/itex] is the angle of projection. Half the height would be [itex] \frac {u^2sin^2(\theta)}{4g} [/itex] . Find the y-component of velocity at half the height . Find the speed by adding the x and y components. Equate it to twice the x-component. Solve for [itex] \theta [/itex] . You should get your answer.
     
  4. May 21, 2016 #3
    X comp : Vicos@t
    Y comp : sqrt(visin@ -2gh/2)
     
  5. May 21, 2016 #4
    X comp : Vicos@ ***
     
  6. May 21, 2016 #5
    Vf = sqrt((vicos@)^2 + Visin@-gh)
     
  7. May 22, 2016 #6
    Are those the correct components?
     
  8. May 22, 2016 #7
    The x-component is correct ([itex] v_icos (\theta) [/itex]) the y-component is wrong. You should learn [itex] \LaTeX [/itex] else your equations are hard to read.
     
  9. May 22, 2016 #8
    Visin@ - gt ?
     
  10. May 23, 2016 #9
    You are just giving me equations, I don't think you are putting in much effort. For the y-component [itex] v_i = v_isin(\theta) [/itex] , acceleration [itex] a = -g [/itex] and height [itex] h = \frac{v_i^2sin^2(\theta)}{4g} [/itex] What equation of motion do you use now to get [itex] v_f [/itex] ?
     
  11. May 23, 2016 #10
    Vf2 = vi2 + 2ax
     
  12. May 23, 2016 #11
    So I will get vi2sin2@/2 for vfy and then i get vf which will be sqrt(vi2cos2@ + vi2sin2@) i equate this to 2
     
  13. May 23, 2016 #12
    2viCos@ which gives me Cos@2 = 1/3 which gives an angle of 54.74 while the answer is 67.8
     
  14. May 23, 2016 #13
    You have clearly made a calculation mistake , using the equation you'll get [itex] v_f^2 = \frac{v_i^2sin^2(\theta)}{2} [/itex] . The resulting equation will be [tex] 4v_i^2cos^2(\theta) = v_i^2cos^2(\theta) + \frac{v_i^2sin^2(\theta)}{2} [/tex] Solve for [itex] \theta [/itex] you'll get your answer.
     
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